A factory produces LED bulbs with an average lifespan of 50,000 hours and a stan
ID: 3323578 • Letter: A
Question
A factory produces LED bulbs with an average lifespan of 50,000 hours and a standard deviation of 4000 hours.
(a) What is the probability that a randomly selected LED bulb lasts for at least 47,500 hours? Draw a sketch of the appropriate region under the normal curve.
(b) In a batch of 100,000 bulbs, how many do you expect to last for less than 45,000 hours?
(c) The top 10% of bulbs last for at least how long?
(d) If it is known that a bulb has already lasted for 52,000 hours, what is the probability that it will last for at least 55,000 hours? 5. In another course (not the one you ar
Explanation / Answer
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 50000
standard Deviation ( sd )= 4000
a.
P(X < 47500) = (47500-50000)/4000
= -2500/4000= -0.625
= P ( Z <-0.625) From Standard Normal Table
= 0.266
P(X > = 47500) = (1 - P(X < 47500)
= 1 - 0.266 = 0.734
b.
P(X < 45000) = (45000-50000)/4000
= -5000/4000= -1.25
= P ( Z <-1.25) From Standard Normal Table
= 0.1056
expected count in a batch of 100,000 bulbs are
= 100000* 0.1056
= 10560
c.
P ( Z > x ) = 0.1
Value of z to the cumulative probability of 0.1 from normal table is 1.281552
P( x-u / (s.d) > x - 50000/4000) = 0.1
That is, ( x - 50000/4000) = 1.281552
--> x = 1.281552 * 4000+50000 = 55126.206262 ~ 55126 hours
d.
If it is known that a bulb has already lasted for 52,000 hours, what is the probability that it will last for at least 55,000 hours
P(X > 52000) = (52000-50000)/4000
= 2000/4000 = 0.5
= P ( Z >0.5) From Standard Normal Table
= 0.3085
P(X > 55000) = (55000-50000)/4000
= 5000/4000 = 1.25
= P ( Z >1.25) From Standard Normal Table
= 0.1056
P( last for >55,000 | already lasted for 52,000 hour)
= P( last for >55,000 and already lasted for 52,000 hour)/ P(already lasted for 52,000 hour)
= P(last for >55,000) / P(already lasted for 52,000 hour)
= 0.1056 / 0.3085
= 0.3423
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